I'm confused about section 9.3 of the book Mirror Symmetry. In particular, I'm confused about the derivation carried out in equations (9.32) to (9.35) where they claim that the partition function is zero if $h'$ has no zeroes.

Specifically, consider a 0-dimensional QFT with fermionic and bosonic variables, defined by the action $$S=S_0 - S_1\psi_1\psi_2 \tag{9.25}$$ with $$S_0(x)=\frac{1}{2}[h'(x)]^2\quad\text{and}\quad S_1=h''(x).\tag{9.28}$$ The partition function is $$Z=\int dxd\psi_1d\psi_2 e^{-S}.\tag{9.26}$$ This action has odd symmetries: $$V_1=\psi_1 \frac{\partial}{\partial x} - h'(x) \frac{\partial}{\partial \psi_2} \quad\text{and}\quad V_2 = \psi_2 \frac{\partial}{\partial x} + h'(x) \frac{\partial}{\partial \psi_1}.\tag{9.30'}$$ This leads to the infinitesimal transformations $$\delta x=\epsilon^1 \psi_1 + \epsilon^2 \psi_2$$ $$\delta \psi_1 = \epsilon^2 h'\tag{9.30}$$ $$\delta \psi_2 = -\epsilon^1 h'.$$ Then, by exploiting these symmetries, they claim that $Z=0$ if $h'$ is everywhere nonzero. This is where my confusion arises.

  1. Is it "legal" to use $$\epsilon^1=\epsilon^2=-\psi_1/h'\tag{1}$$ to change one of the fermionic variables to zero? I thought the $\epsilon$'s had to be infinitesimal for the transformation to be a symmetry and leave the action invariant.

  2. How does this "motivate" the change of variables in equation (9.32) below? $$\hat{x} := x-\frac{\psi_1 \psi_2}{h'}$$ $$\hat{\psi}_1:=\alpha(x) \psi_1\tag{9.32}$$ $$\hat{\psi}_2 := \psi_1 + \psi_2.$$

  3. How did they come up with (9.33) and (9.34)? Equation (9.33) states that $$S(x, \psi_1, \psi_2) = S(\hat{x},0,\hat{\psi}_2)\tag{9.33}$$ presumably because of the symmetry. But what does $S(\hat{x},0,\hat{\psi}_2)$ actually look like? Equation (9.34) is the transformation of the measure from the above change of variables; $$dxd\psi_1d\psi_2 = \left(\alpha(\hat{x}) - \frac{h''(\hat{x})}{(h'(\hat{x}))^2}\hat{\psi}_1 \hat{\psi}_2\right) d\hat{x} d\hat{\psi}_1 d\hat{\psi}_2 .\tag{9.34}$$ I'm not quite sure where this came from.

  4. Equation (9.35) comes directly from plugging in (9.34) (the change in measure) to the partition function. But where is the total derivative in $\hat{X}$ in the second term $$\int e^{-S(\hat{x},0,\hat{\psi}_2)} \frac{h''(\hat{x})}{(h'(\hat{x}))^2} \hat{\psi}_1 \hat{\psi}_2 d\hat{x} d\hat{\psi}_1d\hat{\psi}_2?\tag{9.35'} $$ I guess this may be more apparent if I knew what the function $S(\hat{x},0,\hat{\psi}_2)$ looked like.


1 Answer 1

  1. We should first of all realize that the infinitesimal Grassmann-odd parameters $\epsilon^1$ and $\epsilon^2$ are allowed to depend on the variables $x$, $\psi_1$ and $\psi_2$ in the infinitesimal SUSY transformation (9.30). We clearly will need that in eq. (1).

    OP asks a good question about the status of finite SUSY transformations. To this end let us consider a subclass of infinitesimal parameters of the form $$\begin{align}\epsilon^1(x,\psi_1,\psi_2)~=~&\frac{f^1(x)}{h^{\prime}(x)}\psi_1, \cr \epsilon^2(x,\psi_1,\psi_2)~=~&\frac{f^2(x)}{h^{\prime}(x)}\psi_1, \end{align}\tag{1'} $$ where $f^1(x)$ and $f^2(x)$ are two arbitrary infinitesimal functions. [Here we have used the assumption that $h^{\prime}(x)\neq 0$ for later convenience.] Then the infinitesimal SUSY transformation (9.30) becomes $$ \begin{align}\delta x ~=~& \frac{f^2(x)}{h^{\prime}(x)} \psi_1 \psi_2,\cr \delta \psi_1 ~=~& f^2(x) \psi_1 ,\cr \delta \psi_2 ~=~& -f^1(x) \psi_1 . \end{align}\tag{9.30''} $$ This can be integrated up to the corresponding finite SUSY transformations$^1$ $$ \begin{align} \hat{x} ~=~&x+\frac{F^2(x)}{h^{\prime}(x)}\psi_1\psi_2, \cr \hat{\psi}_1 ~=~& \alpha(x) \psi_1, \qquad\qquad \alpha(x)~:=~1 + F^2(x),\cr \hat{\psi}_2 ~=~&\psi_2 - F^1(x) \psi_1 , \end{align}\tag{9.30'''}$$ where $F^1(x)$ and $F^2(x)$ are two arbitrary finite functions.

    Sketched proof of eq. (9.30'''): The infinitesimal version of eq. (9.30''') is clearly eq. (9.30''). Therefore it is enough to check that the finite SUSY transformation (9.30''') is form-invariant under composition: $$ \begin{align} \hat{\hat{x}} ~=~&\hat{x}+\frac{\hat{F}^2(\hat{x})}{h^{\prime}(\hat{x})}\hat{\psi}_1\hat{\psi}_2 ~=~\hat{x}+\frac{\hat{F}^2(x)}{h^{\prime}(x)}\hat{\psi}_1\psi_2\cr ~=~&x+\frac{F^2(x)+\hat{F}^2(x)\alpha(x)}{h^{\prime}(x)}\psi_1\psi_2~=~x+\frac{\hat{\alpha}(x)\alpha(x)-1}{h^{\prime}(x)}\psi_1\psi_2, \cr \hat{\hat{\psi}}_1 ~=~& \hat{\alpha}(\hat{x}) \hat{\psi}_1 ~=~ \hat{\alpha}(x) \hat{\psi}_1 ~=~ \hat{\alpha}(x) \alpha(x) \psi_1, \cr \hat{\hat{\psi}}_2 ~=~&\hat{\psi}_2 - \hat{F}^1(\hat{x}) \hat{\psi}_1 ~=~\hat{\psi}_2 - \hat{F}^1(x) \hat{\psi}_1\cr ~=~&\psi_2 - \left(F^1(x) - \hat{F}^1(x) \alpha(x)\right) \psi_1 , \end{align}$$ End of sketched proof. $\Box$

    In particular, it is straightforward to check that the action $$S(\hat{x},\hat{\psi}_1,\hat{\psi}_2)~=~S(x,\psi_1,\psi_2)$$ is invariant under the finite SUSY transformation (9.30''').

    It is straightforward to see that the inverse finite SUSY transformation (9.30''') becomes $$ \begin{align} x ~=~&\hat{x}-\frac{F^2(\hat{x})}{\alpha(\hat{x})h^{\prime}(\hat{x})}\hat{\psi}_1\hat{\psi}_2, \cr \psi_1 ~=~& \frac{1}{\alpha(\hat{x})}\hat{\psi}_1, \cr \psi_2 ~=~&\hat{\psi}_2 + \frac{F^1(\hat{x})}{\alpha(\hat{x})}\hat{\psi}_1 .\end{align}\tag{9.30''''}$$ [Hint: Start by establishing the middle equation.]

  2. The ansatz (1') [and the ansatz (1)] were chosen because it is cumbersome (but we suspect not impossible) to integrate the infinitesimal SUSY transformations (9.30) directly. It is much easier to only consider subvariations proportional to $\psi_1$, because then we can repeatedly use the nilpotency $\psi_1^2=0$ to simplify. To arrive at the form (9.32) from eq. (9.30''') now pick $$\begin{align} F^1(x)~=~&F^2(x)~=~-1 , \cr \alpha(x)~:=~&1 + F^2(x) ~=~0,\cr \hat{\psi}_1~=~& \alpha(x) \psi_1~=~0.\end{align}$$ However, this is a singular transformation, so let us first assume that $F^1$ and $F^2$ are $x$-independent constants different from $-1$, and only at the very end of the calculation take the limit going to $-1$. This choice simplifies some algebra as compared to Ref. 1.

  3. Since Ref. 1 identifies Berezin integration with differentiation from right, cf. eq. (9.20), the derivatives in the Jacobian supermatrix are right derivatives. The Jacobian supermatrix becomes $$ \frac{\partial_R(x,\psi_1,\psi_2)}{\partial (\hat{x},\hat{\psi}_1,\hat{\psi}_2)} ~=~\begin{pmatrix} 1+\frac{F^2 h^{\prime\prime}(\hat{x})}{\alpha h^{\prime}(\hat{x})^2}\hat{\psi}_1\hat{\psi}_2 &\ast&\ast \cr 0&\frac{1}{\alpha} &0 \cr 0 & \ast &1 \end{pmatrix}.$$ The superdeterminant/Berezinian becomes $${\rm sdet}\frac{\partial_R(x,\psi_1,\psi_2)}{\partial (\hat{x},\hat{\psi}_1,\hat{\psi}_2)} ~=~\alpha+\frac{F^2 h^{\prime\prime}(\hat{x})}{ h^{\prime}(\hat{x})^2}\hat{\psi}_1\hat{\psi}_2, $$ which explains eq. (9.34).

  4. Finally let's take the limit. The action $$S\left(\hat{x},\hat{\psi}_1\!=\!0,\hat{\psi}_2\right)~=~S_0(\hat{x})~=~\frac{1}{2}h^{\prime}(\hat{x})^2$$ makes the integrand of the second term in (9.35') a total derivative wrt. $\hat{x}$: $$\begin{align} h^{\prime\prime}(\hat{x})~g(h^{\prime}(\hat{x})) ~=~&\frac{d(G(h^{\prime}(\hat{x}))}{d\hat{x}},\cr g(y)~:=~&\frac{e^{-y^2/2}}{y^2},\cr G~:=~&\int\!g~:=~\text{antiderivative of }g. \end{align}$$


  1. K. Hori, S. Katz, A. Klemm, R. Pandharipande, R. Thomas, C. Vafa, R. Vakil, and E. Zaslow, Mirror Symmetry, 2003; Sections 9.2-9.3. The pdf file is available here.
  • 1
    $\begingroup$ I'm going thru the same calculations and I've some doubts about @Qmechanic answer. First when computing the inverse transformation for $\psi_1$, I also obtain a term coming with $\hat{\psi}_1 \hat{\psi}_2$ that I don't see why it should cancel. And for the Jacobian why is the $\alpha(\hat{X})$ not being derivated? $\endgroup$
    – Jasimud
    Jan 17, 2019 at 10:18
  • $\begingroup$ I also can't understand why $\alpha(\hat{X})$ is not differentiated. $\endgroup$
    – Gleeson
    Oct 24, 2023 at 16:25
  • $\begingroup$ Nor why in the fourth point, why we have a total derivative. $\endgroup$
    – Gleeson
    Oct 25, 2023 at 9:36
  • $\begingroup$ Hi @Jasimud and Gleeson. Thanks for the feedback. I updated the answer. $\endgroup$
    – Qmechanic
    Oct 25, 2023 at 14:10
  • $\begingroup$ 1. Thanks, but I still don't see why the $\alpha$ isn't being differentiated. 2. Nor can I understand that total derivative argument. 3. We need some $f(x)$ such that $e^{-\frac{1}{2}(\partial h)^2}\frac{\partial^" h}{(\partial h)^2} = \frac{d}{dx}f(x)$, right? $\endgroup$
    – Gleeson
    Oct 28, 2023 at 13:59

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